What is mean pipe operator between 2 int variants?

Question

Having the follow code -

    enum FileOpenFlags {
    flagREAD = 1, flagWRITE = 2,
    flagCREATE = 4, flagEND = 8,
    flagAPPEND = flagWRITE | flagEND,
    };
    cout << flagAPPEND << endl;

gives 10 . can you explain me what the | did ?

Yes, I can. But I won't as you should have read a C(++) tutorial instead of bugging me with this. — , Aug 28 '12 at 18:15
Those `flag` prefixes can be easily avoided with scoped enums in C++11. — chris, Aug 28 '12 at 18:15
@H2CO3 all my google searches gave results about another meaning of "pipe" , sorry about bugging you . — URL87, Aug 28 '12 at 18:33
That's because it's not a pipe. It's an inclusive or operator. — Pete Becker, Aug 28 '12 at 20:04
@PeteBecker, but some will know it as a pipe *character*. I can easily see where the confusion comes from, and I'm glad to have a place on StackOverflow for Google to find it. — Mark Ransom, Aug 30 '12 at 14:09

score 5 · Accepted Answer · answered Aug 28 '12 at 18:14

5

It did a bitwise or of the two values.

answered Aug 28 '12 at 18:14

Mark Ransom

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score 4 · Answer 2 · answered Aug 28 '12 at 18:15

4

flagWRITE's (2) binary representation is 0010

flagEND's (8) binary representation is 1000

0010 OR 1000 gives you 1010 which equals 10

answered Aug 28 '12 at 18:15

mostar

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score 2 · Answer 3 · answered Aug 28 '12 at 18:15

2

It's called Bitwise OR........

answered Aug 28 '12 at 18:15

Seva Alekseyev

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Ivory_Tower · Answer 4 · 2012-08-28T21:11:59.643

2

It's a bitwise "OR" operator. So the bit value of 2 and 8 respectively get OR'd bitwise.

So:

   1000 (flagEND = 8) 
OR 0010 (flagWRITE = 2)
-----------
 = 1010 (flagAppend = 10)

edited Aug 28 '12 at 21:11

answered Aug 28 '12 at 18:18

Ivory_Tower

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What is mean pipe operator between 2 int variants?

4 Answers4